91Ó°ÊÓ

The vector addition of all available forces acting on any particular object is the net force.

The system's overall net force is given as,

${\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{net}}\mathbf{=}\stackrel{\mathbf{→}}{{\mathbf{F}}_{\mathbf{1}}}\mathbf{+}\stackrel{\mathbf{→}}{{\mathbf{F}}_{\mathbf{2}}}$

Here,$\stackrel{→}{{\mathrm{F}}_1}$is the first external force and $\stackrel{→}{{\mathrm{F}}_1}$is the second external force.

Let insert (40,-70,0)N for ${\stackrel{→}{\mathrm{F}}}_1$and (20,10,0)N for$\stackrel{→}{F_2}$ into the formula of the net force.

$$\begin{gathered} \stackrel{→}{{\mathrm{F}}_{\mathrm{net}}}=\left(40,-70,0\right)\mathrm{N}+\left(20,10,0\right)\mathrm{N} \\ =\left(60,-60,0\right)\mathrm{N} \end{gathered}$$

As a result, the net force on the system is (60,-60,0)N .

The ice's impulse on the hockey puck is,

$∆\stackrel{→}{p}={\stackrel{→}{p}}_f-\stackrel{→}{p_i}$

The end momentum of the hockey puck is$\stackrel{→}{p_f}$ while the starting momentum of the hockey puck is$\stackrel{→}{p_i}$ .

Substitute (0,0,0)kg.m/s for $\stackrel{→}{p_f}$and (0,2,0)kg.m/s for$\stackrel{→}{p_i}$ into the ice’s impulse.

$$\begin{gathered} ∆\stackrel{→}{\mathrm{p}}=\left(0,0,0\right)\mathrm{kg}.\mathrm{m}/\mathrm{s}-\left(0,2,0\right)\mathrm{kg}.\mathrm{m}/\mathrm{s} \\ =\left(0,-2,0\right)\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the ice's impulse acting on the hockey puck is (0,2,0 kg.m/s.

The ice's impulse on the hockey puck is,

$∆\stackrel{→}{p}=\stackrel{→}{F_{Net}}∆t.$

The end momentum of the hockey puck is $\stackrel{→}{p_f}$, the net force operating on the hockey puck is $\stackrel{→}{F_{net}}$, the time is $∆t$, and the initial momentum of the hockey puck is$\stackrel{→}{p_i}$ .

Now rearrange the equation for $\stackrel{→}{F_{Net}}$.

${\stackrel{→}{F}}_{Net}=\frac{∆\stackrel{→}{p}}{∆t}$

Substitute (0,-2,0)kg.m/s for $∆\stackrel{→}{p}$and 3.0 s for$∆t$ into the above formula.

$$\begin{gathered} {\stackrel{→}{\mathrm{F}}}_{\mathrm{Net}}=\frac{\left(0,2,0\right)\mathrm{kg}.\mathrm{m}/\mathrm{s}}{3.0\mathrm{s}}.\left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/\mathrm{s}}\right) \\ =\left(0,-0.667,0\right)\mathrm{N} \end{gathered}$$

Thus, the net force on the hockey puck is$\left(0,-0.667,0\right)\mathrm{N}$.